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zmey [24]
3 years ago
7

Which layer is the igneous rock type?

Physics
2 answers:
Ludmilka [50]3 years ago
6 0

Answer:

option D is correct and Mark me as brainlist

Strike441 [17]3 years ago
5 0
The answer is D that is the only one of the answers that’s right.
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Distance travelled/time taken gives?
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speed

Explanation:

Speed = distance travelled/time taken

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How far will a 600 kg boat travel in 10 s if there is a constant 900 N force on it and it starts from rest?
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here

Explanation:

There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. ... a 3.25-kg object rightward with a constant acceleration of 1.20 m/s/s if the force of ... of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest?

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3 years ago
A deuteron consists of one proton and one neutron. A deuteron moving horizontally enters a uniform, vertical magnetic field of 0
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Answer:

Explanation:

Let the velocity of deuteron be v then force on it in magnetic field

Bqv , B is magnetic field and q is charge on deuteron . This force will provide centripetal force for circular path so

mv² / r = Bqv   m is mass of deuteron and r is radius of circular path

v = Bqr / m

(.5 x 1.6 x 10⁻¹⁹ x 55.6 x 10⁻² )/ 3.34 x 10⁻²⁷

= 13.31 x 10⁶ m /s

6 0
3 years ago
in which of the situations listed below is energy being transferred as heat to the system in order for the system to do work? In
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The answer is option A, i think but i am not sure
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3 years ago
1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc
Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of 3.0 m/s^2.

So, the acceleration of the express train, a=-3 m/s^2.

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, a=-3 m/s^2

So, 0=36+(-3)t

\Rightarrow t= 36/3=12 seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

v^2=u^2+2as

\Rightarrow 0^2=36^2+2(-3)s

\Rightarrow s=\frac{36\times36}{6}

\Rightarrow s=216 meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m  [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0
3 years ago
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